Question

What is the volume of 19.87mol of ammonium chloride NH4CL at STP

Answer 1

$n = \frac{V}{STP} \\ n = \frac{V}{22.4} \\ 19.87 = \frac{V}{22.4} \\ V = 19.87 \times 22.4 \\ V = 445,088 \: l \: of \: NH4CL \:$